00000080.htm

17 COMPACT SETS 63

(3.17.6) (Borel-Lebesgue theorem). In order that a subset of the real
line be relatively compact, a necessary and sufficient condition is that it be
bounded.

In view of (3.17.1), (3.17.4), and (3.17.5), all we have to do is to
prove any closed interval [a, b] is precompact. For each integer /?, let x_k =
a + k(b - a) In (Q^k^n); then the open intervals of center x_k and length
2/n form a covering of [a, b]. Q.E.D.

Remark. If, on the real line, we consider the two distances d_l9 d₂ defined
in Section 3.14, it follows from (3.17.1) that E₂ is not precompact, whereas
EJ is precompact, since the extended real line R, being homeomorphic to the
closed interval [-1, 4-1] of R (3.12), is compact by (3.17.6).

(3.17.7) A necessary and sufficient condition that a subset A of a metric
space E be relatively compact is that every sequence of points of A have a
cluster value in E.

The condition is obviously necessary, by (3.16.1). Conversely, let us
suppose it is satisfied, and let us prove that every sequence (jc_rt) of points of
A has a cluster value in E (which will therefore be in A by (3.13.7)), and hence
that A is compact by (3.16.1). For each /?, it follows from the definition of
closure that there exists y_neA such that d(x_n,y_n)^ l/n. By assumption,
there is a subsequence (y_ni) which converges to a point a; from the triangle
inequality it follows that (x_nk) converges also to a. Q.E.D.

(3.17.8) The union of two relatively compact sets is relatively compact.

From (3.8.8) it follows that we need only prove that the union of two
compact sets A, B is compact. Let (U_A)_A _e _L be an open covering of the sub-
space A u B; each U_A can be written (A u B) n V_A, where V_A is open in E,
by (3.10.1). By assumption, there is a finite subset H (resp. K) of L such
that the subfamily (A nV_A)_A6H(resp. (B n V_;)_AeK) is a covering of A (resp. B).
It is then clear that the family ((A n B) n V_A)_A _eH _w _K is a covering of A u B.

(3.17.9) Let f be a continuous mapping of a metric space E into a metric
space E'. For every compact (resp. relatively compact) subset A of E, /(A)
is compact, hence closed in E' (resp. relatively compact in E').



	17 COMPACT SETS 63 (3.17.6) (Borel-Lebesgue theorem). In order that a subset of the real line be relatively compact, a necessary and sufficient condition is that it be bounded. In view of (3.17.1), (3.17.4), and (3.17.5), all we have to do is to prove any closed interval [a, b] is precompact. For each integer /?, let x_k = a + k(b - a) In (Q^k^n); then the open intervals of center x_k and length 2/n form a covering of [a, b]. Q.E.D.

	Remark. If, on the real line, we consider the two distances d_l9 d₂ defined in Section 3.14, it follows from (3.17.1) that E₂ is not precompact, whereas EJ is precompact, since the extended real line R, being homeomorphic to the closed interval [-1, 4-1] of R (3.12), is compact by (3.17.6).

	(3.17.7) A necessary and sufficient condition that a subset A of a metric space E be relatively compact is that every sequence of points of A have a cluster value in E. The condition is obviously necessary, by (3.16.1). Conversely, let us suppose it is satisfied, and let us prove that every sequence (jc_rt) of points of A has a cluster value in E (which will therefore be in A by (3.13.7)), and hence that A is compact by (3.16.1). For each /?, it follows from the definition of closure that there exists y_neA such that d(x_n,y_n)^ l/n. By assumption, there is a subsequence (y_ni) which converges to a point a; from the triangle inequality it follows that (x_nk) converges also to a. Q.E.D.

	(3.17.8) The union of two relatively compact sets is relatively compact. From (3.8.8) it follows that we need only prove that the union of two compact sets A, B is compact. Let (U_A)_A _e _L be an open covering of the sub- space A u B; each U_A can be written (A u B) n V_A, where V_A is open in E, by (3.10.1). By assumption, there is a finite subset H (resp. K) of L such that the subfamily (A nV_A)_A6H(resp. (B n V_;)_AeK) is a covering of A (resp. B). It is then clear that the family ((A n B) n V_A)_A _eH _w _K is a covering of A u B.

	(3.17.9) Let f be a continuous mapping of a metric space E into a metric space E'. For every compact (resp. relatively compact) subset A of E, /(A) is compact, hence closed in E' (resp. relatively compact in E').